How do you solve the equation #ax^2+5=3#?

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Answer 1 · 2018-05-18T22:36:46

In terns of x:

#x = +-sqrt((-2)/a)#

if a is not negative the solution will be imaginary.

#ax^2 + 5 = 3#

#ax^2 + 5 -5 = 3 - 5#

#ax^2 = -2#

#(ax^2)/a = (-2)/a#

#x^2 = (-2)/a#

#sqrt(x^2) = +-sqrt(-2/a)#

#x = +-sqrt((-2)/a)#